Math4201 Topology II (Lecture 12)

Algebraic topology

Fundamental group

Recall from last lecture, the $(\Pi_1(X,x_0),*)$ is a group, and for any two points $x_0,x_1\in X$, the group $(\Pi_1(X,x_0),*)$ is isomorphic to $(\Pi_1(X,x_1),*)$ if $x_0,x_1$ is path connected.

Definition of simply connected

A space $X$ is simply connected if

• $X$ is path-connected  ($\forall x_0,x_1\in X$, there exists a continuous function $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$)
• $\Pi_1(X,x_0)$ is the trivial group for some $x_0\in X$

Lemma for simply connected space

In a simply connected space $X$, and two paths having the same initial and final points are path homotopic.

Definition of group homomorphism induced by continuous map

Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.

$h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.

Theorem composite of group homomorphism

If $h:(X,x_0)\to (Y,y_0)$ and $k:(Y,y_0)\to (Z,z_0)$ are continuous maps, then $k_* \circ h_*:\Pi_1(X,x_0)\to \Pi_1(Z,z_0)$ where $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$, $k_*:\Pi_1(Y,y_0)\to \Pi_1(Z,z_0)$,is a group homomorphism.

Corollary of composite of group homomorphism

Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:\Pi_1(X,x_0)\to \Pi_1(X,x_0)$.

If $h$ is a homeomorphism with the inverse $k$, with

This induced $h_*: \Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ is an isomorphism.

Corollary for homotopy and group homomorphism

If $h,k:(X,x_0)\to (Y,y_0)$ are homotopic maps form $X$ to $Y$ such that the homotopy $H_t(x_0)=y_0,\forall t\in I$, then $h_*=k_*$.